A*搜索求解8数码问题python实现
时间: 2023-10-08 22:10:14 浏览: 115
好的,这是一个比较经典的算法案例,下面是 A* 搜索求解 8 数码问题的 Python 实现:
```python
import heapq
class Puzzle:
def __init__(self, start, goal):
self.start = start
self.goal = goal
def manhattan_distance(self, board):
distance = 0
for i in range(3):
for j in range(3):
if board[i][j] == 0:
continue
x, y = divmod(self.goal.index(board[i][j]), 3)
distance += abs(i - x) + abs(j - y)
return distance
def solve(self):
opened = []
heapq.heappush(opened, (self.manhattan_distance(self.start), self.start, self.start.index(0), 0, []))
closed = set()
directions = ['Up', 'Down', 'Left', 'Right']
while opened:
_, board, zero_index, count, path = heapq.heappop(opened)
if board == self.goal:
return path
closed.add(board)
for direction in directions:
if direction == 'Up' and zero_index > 2:
new_board = board[:]
new_board[zero_index], new_board[zero_index - 3] = new_board[zero_index - 3], new_board[zero_index]
if tuple(new_board) not in closed:
new_path = path[:]
new_path.append('U')
heapq.heappush(opened, (self.manhattan_distance(new_board) + count + 1, new_board, zero_index - 3, count + 1, new_path))
elif direction == 'Down' and zero_index < 6:
new_board = board[:]
new_board[zero_index], new_board[zero_index + 3] = new_board[zero_index + 3], new_board[zero_index]
if tuple(new_board) not in closed:
new_path = path[:]
new_path.append('D')
heapq.heappush(opened, (self.manhattan_distance(new_board) + count + 1, new_board, zero_index + 3, count + 1, new_path))
elif direction == 'Left' and zero_index not in (0, 3, 6):
new_board = board[:]
new_board[zero_index], new_board[zero_index - 1] = new_board[zero_index - 1], new_board[zero_index]
if tuple(new_board) not in closed:
new_path = path[:]
new_path.append('L')
heapq.heappush(opened, (self.manhattan_distance(new_board) + count + 1, new_board, zero_index - 1, count + 1, new_path))
elif direction == 'Right' and zero_index not in (2, 5, 8):
new_board = board[:]
new_board[zero_index], new_board[zero_index + 1] = new_board[zero_index + 1], new_board[zero_index]
if tuple(new_board) not in closed:
new_path = path[:]
new_path.append('R')
heapq.heappush(opened, (self.manhattan_distance(new_board) + count + 1, new_board, zero_index + 1, count + 1, new_path))
return None
puzzle = Puzzle([1, 2, 3, 4, 5, 6, 0, 7, 8], [1, 2, 3, 4, 5, 6, 7, 8, 0])
print(puzzle.solve())
```
这段代码实现了 A* 搜索求解 8 数码问题,使用了 Manhattan 距离作为启发函数,使用堆来进行优先队列的实现。